Question

The GPAs of all students enrolled at a large university have an approximately normal distribution with a mean of 3.02 and a standard deviation of .29.Find the probability that the mean GPA of a random sample of 20 students selected from this university is 3.10 or higher.

Answer 1

Answer:

10.93% probability that the mean GPA of a random sample of 20 students selected from this university is 3.10 or higher.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 3.02, \sigma = 0.29, n = 20, s = \frac{0.29}{\sqrt{20}} = 0.0648$

Find the probability that the mean GPA of a random sample of 20 students selected from this university is 3.10 or higher.

This is 1 subtracted by the palue of Z when X = 3.10. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{3.1 - 3.02}{0.0648}$

$Z = 1.23$

$Z = 1.23$ has a pvalue of 0.8907

1 - 0.8907 = 0.1093

10.93% probability that the mean GPA of a random sample of 20 students selected from this university is 3.10 or higher.